Q. 55 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 51–56, solve for x, 

$\left|\begin{array}{rrr}x& 2& 3\\ 1& x& 0\\ 6& 1& -2\end{array}\right|=7$

Step-by-Step Solution

Answer

The value of x is $0 \text{or }9$.

Step 1. Given Information

We are given, 

$\left|\begin{array}{rrr}x& 2& 3\\ 1& x& 0\\ 6& 1& -2\end{array}\right|=7$

We have to find the value of x.

Step 2. Solving the determinant

Solving the determinant, we get 

$$\begin{gathered} \left|\begin{array}{rrr}x& 2& 3\\ 1& x& 0\\ 6& 1& -2\end{array}\right|=7 \\ x\left|\begin{array}{cc}x& 0\\ 1& -2\end{array}\right|-2\left|\begin{array}{cc}1& 0\\ 6& -2\end{array}\right|+3\left|\begin{array}{ll}1& x\\ 6& 1\end{array}\right|=7 \\ x\left[\right(x\left)\right(-2)-(0\left)\right(1\left)\right]-2\left[\right(1\left)\right(-2)-(0\left)\right(6\left)\right]+3\left[\right(1\left)\right(1)-(x\left)\right(6\left)\right]=7 \\ x(-2 x)-2(-2)+3(1-6 x)=7 \\ -2x^2-18x+7=7 \\ -2x^2-18x=0 \\ 2 x(x-9)=0 \\ 2x=0 \\ x=0 \end{gathered}$$

or

$$\begin{gathered} x-9=0 \\ x=9 \end{gathered}$$